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MathematicsHigh schoolHard

Local extrema of a cubic function

Find the increasing and decreasing intervals and the local extrema of the function

f(x)=x3−3x.f(x) = x^3 - 3x.
Hintshow

A function is increasing where f′(x)>0f'(x) > 0 and decreasing where f′(x)<0f'(x) < 0. An extremum occurs at a point where the derivative changes sign.

Solutiontry it on your own first

Step 1. Find the domain

The function f(x)=x3−3xf(x) = x^3 - 3x is a polynomial, so it is defined and differentiable for every real number:

D=RD = \mathbb{R}

Step 2. Find the derivative

Differentiate each term separately:

(x3)′=3x2,(3x)′=3(x^3)' = 3x^2, \qquad (3x)' = 3

So:

f′(x)=3x2−3\boxed{f'(x) = 3x^2 - 3}

Step 3. Find where the derivative is zero

Solve the equation:

3x2−3=03x^2 - 3 = 0

Divide both sides by 3:

x2−1=0x^2 - 1 = 0

Factorise the left-hand side as a difference of two squares, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b):

(x−1)(x+1)=0(x - 1)(x + 1) = 0

A product is zero when one of its factors is zero:

x=−1orx=1\boxed{x = -1 \quad \text{or} \quad x = 1}

Step 4. Find the sign of the derivative

Write the derivative in factorised form:

f′(x)=3(x−1)(x+1)f'(x) = 3(x - 1)(x + 1)

Its graph is a parabola that opens upward (the coefficient of x2x^2 is 3>03 > 0) with zeros at −1-1 and 11. So it is positive outside the zeros and negative between them. Check this with three sample numbers:

f′(−2)=3⋅4−3=9>0f′(0)=−3<0f′(2)=3⋅4−3=9>0\begin{aligned} f'(-2) &= 3 \cdot 4 - 3 = 9 > 0 \\ f'(0) &= -3 < 0 \\ f'(2) &= 3 \cdot 4 - 3 = 9 > 0 \end{aligned}

Put the results in a table:

xx sign of f′(x)f'(x) f(x)f(x)
(−∞,−1)(-\infty, -1) ++ increasing
−1-1 00 maximum
(−1,1)(-1, 1) −- decreasing
11 00 minimum
(1,∞)(1, \infty) ++ increasing

Step 5. Read off where the function increases and decreases

Where f′(x)>0f'(x) > 0 the function is increasing, and where f′(x)<0f'(x) < 0 it is decreasing. The function is continuous, so the endpoints can be included. The function ff:

  • is increasing on (−∞,−1](-\infty, -1] and on [1,∞)[1, \infty),
  • is decreasing on [−1,1][-1, 1].

Step 6. Find the extrema and their values

At x=−1x = -1 the derivative changes sign from ++ to −-, so the function has a local maximum there. Its value:

f(−1)=(−1)3−3⋅(−1)=−1+3=2\begin{aligned} f(-1) &= (-1)^3 - 3 \cdot (-1) \\ &= -1 + 3 \\ &= 2 \end{aligned}

At x=1x = 1 the derivative changes sign from −- to ++, so the function has a local minimum there:

f(1)=13−3⋅1=1−3=−2\begin{aligned} f(1) &= 1^3 - 3 \cdot 1 \\ &= 1 - 3 \\ &= -2 \end{aligned}

Answer

fmax⁡=f(−1)=2\boxed{f_{\max} = f(-1) = 2}
fmin⁡=f(1)=−2\boxed{f_{\min} = f(1) = -2}

The function is increasing on (−∞,−1](-\infty, -1] and on [1,∞)[1, \infty) and decreasing on [−1,1][-1, 1]. It has a local maximum of 22 at x=−1x = -1 and a local minimum of −2-2 at x=1x = 1.

The theory behind this exerciseCourse: Derivatives from scratch

Sources

We checked the facts on this page against these sources. All of them may be reused freely: they carry an open licence or are not covered by copyright.

  1. Active Calculus (Boelkins) - 3.3 Using derivatives to identify extreme values (critical numbers, first derivative test, sign charts) · CC BY-SA 4.0
  2. Dale Hoffman - Contemporary Calculus, 3.1 Finding Maximums and Minimums (PDF) · CC BY 3.0 US
  3. Dale Hoffman - Contemporary Calculus, 3.3 The First Derivative and the Shape of f (increasing and decreasing, first derivative test) (PDF) · CC BY 3.0 US
  4. Calaway, Hoffman and Lippman - Applied Calculus, 2.7 Optimization (critical numbers, first derivative test for extremes) (PDF) · CC BY 3.0 US
  5. Wikiversity - Mathematics for Applied Sciences (Osnabrück 2023-2024), Part I, Lecture 15 (local extrema need f' = 0; the sign of f' gives where f increases or decreases) · CC BY-SA 4.0
  6. Wikibooks - Math for Non-Geeks, Derivative and local extrema (sign change of the derivative) · CC BY-SA 4.0

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